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Re: PS: Probability [#permalink]
24 Jul 2008, 09:56
I get B, 3/8.
There are 96 total possibilities (i.e, the denominator).
There are 12 numbers 96/8 that are multiples of/divisible by 8 themselves.
If n = 6, then you have 6*(6+1)(6+2) or 6*7*8, so that number will be divisible by 8. Same for n = 14. 14(14+1)(14+2), 14*15*16...also divisible by 8 because 16 is divisible by 8. For every number that is divisible by 8, (every 8th number to 96) will have 3 values for n that create a number with n(n+1)(n+2) that will be divisible by 8, so I multiplied 12 by 3 = 36/96
36/96 = 6 / 16 = 3 / 8
EDIT: I just did this in Excel and came up with 5/8 (60/96). I went wrong because I didn't consider that n=2 will have 2*3*4, or 8*3, which will be divisible by 8 also!
What's the best way to figure this out?
My answer to my own question:
We know that any set of number n(n+1)(n+2) that will be divisible by 2, three separate times ( like 16 /2 = 8, 8/2 = 4, 4/2 = 2 is divisible by 2 three times) will be divisible by 8. Since we know that 2 + 3 + 4 works 2 /2 , and 4/ 2, = 2 ...2/2 = 3 times... every even number will satisfy this question, so 48 of them.
Then we also know that any odd number that includes a multiple of 8 as (n+1) [ n+2 doesn't matter be odd + 2 = odd and will never be divisble by 8] will also satisfy this. There is only 1 odd number that satisfies this for every multiple of 8 up to 96. We've already established there 96/8 = 12, so there are 12 odd numbers that also satisfy this question. so 48 + 12 = 60. Probability = 60/96 or 5/8
tarek99 wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8?
a) 1/4 b) 3/8 c) 1/2 d) 5/8 e) 3/4
Any fast way to solve this???
_________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
Re: PS: Probability [#permalink]
24 Jul 2008, 10:43
droopy57 wrote:
tarek99 wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8?
a) 1/4 b) 3/8 c) 1/2 d) 5/8 e) 3/4
Any fast way to solve this???
D, what is difficulty of this problem? took me about 3-4 minutes to solve. I broke down #s but I think there is a simpler way.
(1-10), 4,5,6,7,8,10 were the only ones that worked.
6*10= 60/96
Found it wasnt as difficult as it was finding out which digits worked...
First, 5 does not work. 5*6*7 = 210 210 / 8 = 26.25.
Second, 2 does work. 2*3*4 = 24, 24/8 = 3.
Also, where did you get " *10 " ? if you have 6 possible for 1-10, and you are saying you have ten "groups" of 10, that's incorrect. The numbers go from 1 - 96. There can't possibly be 10 groups of 10 in 96 numbers. _________________
------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.
Re: PS: Probability [#permalink]
24 Jul 2008, 11:23
jallenmorris wrote:
droopy57 wrote:
tarek99 wrote:
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n+1)(n+2) will be divisible by 8?
a) 1/4 b) 3/8 c) 1/2 d) 5/8 e) 3/4
Any fast way to solve this???
D, what is difficulty of this problem? took me about 3-4 minutes to solve. I broke down #s but I think there is a simpler way.
(1-10), 4,5,6,7,8,10 were the only ones that worked.
6*10= 60/96
Found it wasnt as difficult as it was finding out which digits worked...
First, 5 does not work. 5*6*7 = 210 210 / 8 = 26.25.
Second, 2 does work. 2*3*4 = 24, 24/8 = 3.
Also, where did you get " *10 " ? if you have 6 possible for 1-10, and you are saying you have ten "groups" of 10, that's incorrect. The numbers go from 1 - 96. There can't possibly be 10 groups of 10 in 96 numbers.
At second glance, I have no idea how I came up with that #. Pure luck.
Attacking this problem again and having major issues.... There is a pattern but taking to long to uncover.
Re: PS: Probability [#permalink]
24 Jul 2008, 12:47
droopy57 wrote:
At second glance, I have no idea how I came up with that #. Pure luck.
Attacking this problem again and having major issues.... There is a pattern but taking to long to uncover.
help?
gmatnub had it right -- just didn't explain it fully.
all even numbers should qualify because the product of consecutive even numbers is always divisible by 8. This is because one of the consecutive even numbers will be divisible by 2 and the other will be divisible by 4.
that gets us to 48 of the 96 numbers.
now, you have to count the odd numbers such that n+1 is divisible by 8. This is the case with all x-1 such that x is divisible by 8.
96/8 = 12 numbers b/w 1 and 96.
(48+12)/96 = 5/8
gmatclubot
Re: PS: Probability
[#permalink]
24 Jul 2008, 12:47